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anonymous

  • 5 years ago

Hello!! Is anyone good with evaluating indefinite integrals?

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  1. anonymous
    • 5 years ago
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    This one seems easy but I could sure use confirmation on the correct answer!! Lol.

  2. anonymous
    • 5 years ago
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    post the question

  3. anonymous
    • 5 years ago
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    Posting!!

  4. anonymous
    • 5 years ago
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    uh uhs, amistre is here :P

  5. amistre64
    • 5 years ago
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    .......TRIG......

  6. anonymous
    • 5 years ago
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    i meant uh ohs*

  7. anonymous
    • 5 years ago
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    I couldn't figure out how to post a division with out the sign!

  8. amistre64
    • 5 years ago
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    give me a T; give me an R; give me an IG...whats that spell? trouble?

  9. anonymous
    • 5 years ago
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    lols nah, spells awesomeness

  10. amistre64
    • 5 years ago
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    post it your best and well decipher it :)

  11. anonymous
    • 5 years ago
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    \frac{numer}{deno}

  12. anonymous
    • 5 years ago
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    ArcsinX + C

  13. anonymous
    • 5 years ago
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    ^-- in the equation editor and like amistre said it's not a big deal

  14. anonymous
    • 5 years ago
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    lols all yours amistre

  15. amistre64
    • 5 years ago
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    integrate arcsin + C? or is that your answer that yougot?

  16. anonymous
    • 5 years ago
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    That's my answer - I just have to evaluate the integral

  17. amistre64
    • 5 years ago
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    whats the integral so that we can pervue its intricate innards :)

  18. anonymous
    • 5 years ago
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    \[\frac{1}{\sqrt{1 - x^2}}\] is this the integral?

  19. anonymous
    • 5 years ago
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    without the integral sign...or the dx...

  20. anonymous
    • 5 years ago
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    \frac\[\int\limits_{?}^{?}\frac{1}{1+x^2} dx\]

  21. amistre64
    • 5 years ago
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    thats a arctan..

  22. anonymous
    • 5 years ago
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    yup, amistre's right

  23. amistre64
    • 5 years ago
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    i saw it in a movie lol

  24. anonymous
    • 5 years ago
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    lols "down the rabbit hole"?

  25. amistre64
    • 5 years ago
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    :) tan^1(x) + C...are there any bounds to evaluate?

  26. amistre64
    • 5 years ago
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    ..^-1..

  27. anonymous
    • 5 years ago
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    \[[\int\limits] dx 1 + x2 = \arctan x + C \]

  28. anonymous
    • 5 years ago
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    Was that all I needed to refer to?! I have it in my notes but the 'dx' was in the num

  29. amistre64
    • 5 years ago
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    its like learning greek all over again ;)

  30. anonymous
    • 5 years ago
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    something like this? \[\int\limits_{}^{}\frac{1}{x^2 + 1}dx\]

  31. anonymous
    • 5 years ago
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    Noooo kidding! lol

  32. anonymous
    • 5 years ago
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    Noooo kidding! lol

  33. amistre64
    • 5 years ago
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    \[\int\limits_{} \frac{1}{1+x^2} dx \rightarrow \tan^{-1}(x) + C\]

  34. anonymous
    • 5 years ago
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    for answering this question, amistre...i give you...a MEDAL

  35. anonymous
    • 5 years ago
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    -applause-

  36. amistre64
    • 5 years ago
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    yay!! ..... its prolly foiled covered chocolate...so I accept :)

  37. anonymous
    • 5 years ago
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    lols how'd you guess? enjoy! cheers

  38. anonymous
    • 5 years ago
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    Will arctan x + c be correct?

  39. anonymous
    • 5 years ago
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    -nod-

  40. amistre64
    • 5 years ago
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    as long as you havent left anyting out, yes; but arctan is a dated term

  41. amistre64
    • 5 years ago
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    i would know; im dated ;)

  42. anonymous
    • 5 years ago
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    lol!

  43. anonymous
    • 5 years ago
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    Can you guys help me with another problem/confirm my answer being right or wrong?!!

  44. amistre64
    • 5 years ago
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    perhaps, but display the question first; just makes it easier :)

  45. anonymous
    • 5 years ago
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    lol - kk!!

  46. anonymous
    • 5 years ago
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    Eval the Indefinite Integral :\[\int\limits\limits_{?}^{?}\frac{x^3 dx}{\sqrt[4]{5x^4-1}}\]

  47. amistre64
    • 5 years ago
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    you want a 15 for that :)

  48. anonymous
    • 5 years ago
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    I got: 1/15 (5x^4)^3/4 + C

  49. amistre64
    • 5 years ago
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    its refered to as "u- substitution" u = 5x^4 - 1; du = 20x^3 dx dx = du/20x^3

  50. anonymous
    • 5 years ago
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    \[\frac{1}{15} (5x^{4})^{3/4} + C\]

  51. amistre64
    • 5 years ago
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    \[\int\limits_{} \frac{x^3 }{20x^3 u^{1/4}}du\]

  52. amistre64
    • 5 years ago
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    i mistyped 15; meant 20 :) the x^3s cancel; pull out the 1/20; and integrate the u^(-1/4)

  53. amistre64
    • 5 years ago
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    \[\frac{1}{20} \int\limits_{} u^{-1/4} du \rightarrow \frac{1}{20} \frac{u^{(-1+ 4)/4}}{(-1+4)/4}\]

  54. amistre64
    • 5 years ago
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    u^(3/4)/(3/4) 4 * 4root(u^3) ------------ 20 * 3 4root(u^3) ------------ = 4root[5x^4-1]/15 5 * 3

  55. amistre64
    • 5 years ago
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    forgot to ^3 my (5x^4 - 1)

  56. amistre64
    • 5 years ago
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    \[\frac{\sqrt[4]{(5x^4 -1)^3}}{15} +C\]

  57. anonymous
    • 5 years ago
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    You are great!!!!! So final answer: \[\frac{1}{20} (5x^{4} -1)^{3/4} + C?\]

  58. amistre64
    • 5 years ago
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    that looks comparabley good :)

  59. amistre64
    • 5 years ago
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    ack!!...1/15.....not 1/20

  60. anonymous
    • 5 years ago
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    ahh!!

  61. anonymous
    • 5 years ago
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    Fantastic!! lol - I almost messed that one up! Thank you so much! I have another problem I might post a new question so I can award medals - lol

  62. amistre64
    • 5 years ago
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    im sure we will be around :)

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